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Friday, August 27, 2010

I know what you’re thinking…”Wow! This guy must be really smart because he designed a switch mode power supply for his project.” Well, I hate to disappoint, but I’m not really that smart at electronics (that’s why I’m doing this project…to learn!)

The truth is, I was looking at building my own SMPS, but it looked like a chore even considering there are some really easy to use chips like the LM2575. Picky PCB layout problems and the fact that I don’t own an inductor of any kind, really made me think twice about trying. Oh, I designed a SMPS in Eagle, but I have yet to try it because I came up with a…how should I put this…dumb simple, solution. That is, simple enough I can do it and it just works. No fussing around with it.

First, I knew that I needed something to convert from my 11.1V (call it 12V) LiPo battery down to a useful 3.3V. Looking at typical linear voltage regulators, I could see that I would be burning up lots of my battery as heat. In fact, a 12V to 3.3V linear regulator is only about 28% efficient. That’s a deal breaker. This scoreboard needs to be as efficient as possible in the power department because it could be used for hours at a time. Incidentally, during development I was using a 5V LDO regulator and just converting from 12V to 5V caused the regulator to heat up to where it will burn you if you touch it. I don’t want that in my scoreboard box. The LEDs are going to be warm enough. My other option was to look at a SMPS.

Looking up SMPS online, I found that they tend to be about 80 to 90% efficient in the voltage and current ranges that I am expecting. That’s a HUGE improvement. This is when I made the decision to do a SMPS or bust. Little did I know that I actually had a SMPS lurking in my desk drawer that would do the job perfectly. While I was pondering SMPS datasheets, I noticed a car charger from a long defunct cell phone laying on my desk. “Hey!” I said to myself, “that thing converts from 12V car voltage to 5V USB!” My search for a SMPS was over.

Gutting the charger, I found a small circuit with a couple resistors, a diode, an 8 pin unmarked dip, and (the tell tale sign of a SMPS), an inductor. But this SMPS outputs 5V, not 3.3V, which is what I need. Having done a bit of research already, I knew that the voltage on a variable SMPS chip is set with a potential divider. So I set out to recreate the schematic of this simple board. Soon, I had identified the two resistors that create the voltage divider, figured out which one I needed to replace, calculated the new resistor value, and replaced the resistor with an appropriate value for 3.3V output.

Honestly, I didn’t think it could be this simple, but apparently it was. When I powered on the circuit, it measured exactly 3.295V. Sweet!! I had concerns that perhaps this SMPS couldn’t handle the current required for running all my electronics, but now that I have it hooked up I can see that it’s not going to be an issue. In a steady state, my electronics (minus the digits which run on 12V anyway) take anywhere from 30 to 50mA with peaks up to 200mA. The SMPS can handle that, no problem, if it ever charged a phone.

Here’s a pic of my repurposed SMPS.

For those of you that are impeccably observant, you may have noticed that there are two resistors connected together. That’s just me not being able to find a 1000 ohm resistor, so I made one with two 500 ohm resistors. Yeah, okay, it’s not that clean, but it works and I’m all about things that just work.

That’s all for now!

Edit:

I've had a couple comments from folks clarifying the efficiency assumptions that I made in this post. This inspired me to measure the actual efficiency of the supply in question. Here are my results:

I attached a 9v source to the input and a purely resistive load of 58 Ohms on the output.

Vin = 9v
Vout = 3.3v
Iin(no load) = 3.6 mA
Iin = 30 mA
Iout = 55 mA

Time to whip out Ohm's Law. As you can see, the supply consumes 3.6 mA current with no load connected. Overall, the supply is

On a positive note, the current that the supply consumes remains constant, so the higher the current, the higher the efficiency (to a point). At higher current, I suppose this supply could easily reach 75% efficiency.

6 comments:

Any linear regulator should have an efficiency of the ratio between the desired voltage and the input voltage. For example, 3.3/12 = 0.275 or 27.5%. You are right on with 28%. But with your battery actually being 11.1v, the efficiency would be ~29.7%, of course varying with battery depletion.

Fair enough, however I do say at the beginning "call it 12V". This is primarily because an 11.1V LiPo pack is well over 12V when fully charged and remains so for quite a long time. I think it's a valid generalization in lieu of an overly scientific (for me) graph of voltage and efficiency over time.

And a big thanks and congratulations for being the first commentor on my website. Cheers!!

Nice job!However, efficiency should be calculated in watts not in volts. I don't think your first linear regulator had a efficiency of 28% approx, it's low. You should consider calculating it through this formula and tell us what the result is:

Power IN = Power OUT + Losses in Regulator

Power IN = Vbatt * Ibatt

Power OUT = Vregulated * I

Losses = (Vbatt - Vregulated) * I

"I" equals the mA that your circuit is "sucking" from the regulator.

So,

Efficiency = Power out / Power in

which means that the hotter your regulator gets the less efficient it is.

Very nice, I am in the same boat with you. I understand many electronic basics, but am not comfortable enough to design a circuit like this, but I'm pretty sure I've got 1 or 2 of these kickin around. Will have to give this a shotThanks for the idea

Thanks for your comment. I guess I kind of glossed over how I calculated the efficiency of of the LDO, but honestly that wasn't really the point of the post. You are correct in that you use input and output power as the common denominator for calculating efficiency. In the case of an LDO regulator, it is close enough to assume the current to be the same on both the input and the output, thus causing current to cancel out of the efficiency equation.